Information for reliability block diagrams often comes from a
mixture of sources, including in-house testing,
manufacturer data or reliability prediction methods. For
any component with information that comes from a manufacturer or
a reliability prediction, only the parameters (e.g. MTBF
for an exponential distribution or β and
η for a Weibull distribution) are available for
that component. Without the actual failure times that
were used to compute these parameters, it is impossible
to compute the variance of reliability for that
component.

In order to compute bounds on system
reliability, it is necessary to know the expected value
and the variance of reliability for each component in
the system, so the bounds on system
reliability are not available in these cases. The bounds
on system reliability can be obtained, however, for the special case
where all the times-to-failure data for all the components
in the system are available. There are several
methods to compute these bounds. In this article we will
use a simple example to illustrate the method used in Weibull++.

The basic steps to calculate bounds on system
reliability are as follows:

Obtain the parameters and the variances and
covariances of those parameters for each component
in the system.

Compute the variance in reliability for each
component in the system.

Obtain an expression for the variance in system
reliability.

Use the logit transform to obtain the bounds on
system reliability.

ExampleOn the day before Julie’s boss went on
vacation, he assigned her a new project. He wanted Julie
to take the subsystem test data that had just come back
from the lab and figure out what the 95% one-sided lower
confidence bound on system reliability would be at 150
hours. Julie knew that she could use the reliability
block diagram feature in Weibull++ to compute this information
quickly, but she was curious about how these values were
calculated. Since she had a very simple system, she
decided to try to do the calculations step-by-step and
compare them to the results of the block diagram.

Step 1: Table 1 shows the times-to-failure for both subsystems.

Table 1: Times-to-Failure for
Subsystems 1 and 2

Subsystem 1

Subsystem 2

72

171

313

3

204

106

187

316

5

234

110

214

335

25

239

134

218

346

61

257

147

225

386

71

316

150

245

426

97

634

160

262

149

161

306

186

Julie
pasted the data into two Data Sheets in a single
Standard Folio in Weibull++. She calculated the Weibull
distribution parameters for each data set as shown in
Figures 1 and 2. She found that for subsystem 1,
β =
2.5663 and
η = 254.8755 hours, and for subsystem 2,
β =
0.7605 and
η = 183.5255 hours.

Figure 1: Data for
Subsystem 1

Figure 2: Data for Subsystem 2

Next, Julie found the reliability for each subsystem at
150 hours using the QCP, as shown in Figure 3.

Figure 3: Reliability for Subsystems 1 and 2 at
150 Hours

Finally, Julie needed to obtain the
variances and covariances of the parameters for each
subsystem. These values are estimated from the inverse
of the local Fisher matrix. (For more information about
the Fisher matrix, please see the Confidence Bounds for
Competing Failure Modes section at
http://reliawiki.org/index.php/Competing_Failure_Modes_Analysis).
She clicked the Show Fisher Matrix button on the
Confidence Bounds page of the QCP to obtain the
following:

Table 2: Fisher Matrices for Subsystems 1
and 2

Subsystem 1

Subsystem 2

Var(β)=0.1845

Cov(β,η)=3.2355

Var(β)=0.0307

Cov(β,η)=0.6821

Cov(η,β)=3.2355

Var(η)=497.8763

Cov(η,β)=0.6821

Var(η)=5420.7728

Step 2: At this point, Julie dusted
off her old statistics book to help her determine the
variance of reliability for each component. She turned
to the page that had the formula for the variance of a
function of multiple variables. In general:

For the
Weibull distribution:

Since the Weibull reliability
is:

Julie was able to compute the partial derivatives
as follows:

Substituting these into the expression
for variance yields:

Using the values she obtained in
Step 1, Julie computed the variance of reliability at
150 hours for each subsystem. She found that for
subsystem 1, Var(R) = 0.0053922 and for subsystem 2, Var(R)
= 0.0126361.

Step 3: The subsystems are arranged
reliability-wise in parallel, so the equation for the
reliability of the system is:

Once again, Julie
turned to the formula for the variance of a function of
multiple variables. Since the component reliabilities
are independent, the covariance terms are zero.
Therefore, for her system:

Step 4: Julie knew that the logit
transform is used to find the bounds on reliability
given an expected value and variance of reliability.
Therefore, the expression for the lower bound of system
reliability is given by:

where Kα is defined as:

For a one-sided bound,
α = (1 – confidence level), so
she used
α = 0.05 as the input for the Quick Statistical
Reference to determine the value of Kα as shown in
Figure 4.

Figure 4: Calculation of Kα for a 95%
One-Sided Lower Bound

Julie now had everything she
needed to compute the lower bound of the system
reliability. Plugging the values into the equation for
the lower bound of the system reliability at a 95%
confidence at 150 hours, Julie found that the lower
bound on reliability was 76.53%.

Verification
Julie created a new block diagram with her subsystems in
parallel, as shown in Figure 5.

Figure 5: Block
Diagram for Parallel Subsystems

She calculated the
diagram and opened the QCP. She set up a 95% lower
one-sided bound and computed the reliability of the
system as shown in Figure 6. Julie was pleased to find
that her calculations matched those of Weibull++, and
confident that she understood how confidence bounds on
system reliability are obtained.